Crazy Cubology

Ok, apologies for the long absence. I have news from the front - the war is going well.
No, wait, that was some other thing...

Well anyways, I think I have managed to strengthen my conjectures about what a Rubik's cube really is, and so here they are along with some ideas.

Conjecture #1: The set of puzzles, which includes any that have a general Platonic form and are deeply sliced, are general models of switching theory and computation.

Idea: you can buy kitsets of the puzzles, the most popular or available is for the 3x3x3. If you construct only the black plastic components and ignore the colored ones (stay colorblind) what do you have? Could this construction be done wearing a blindfold?

Yes to the last question - you know how, or can learn, to put a disassembled cube together. The other question is open to interpretation; I would say what you have with an assembled, uncolored Rubik's cube is a breadboard or a bench-test frame. What you test is conjectures about what using the colored stickers will change, in terms of locating any element or group of elements.

You conjecture that, given squares of black plastic, you can assemble the trihedral and dihedral elements in black (and white). The inner facets of these elements are not squares of plastic though.

At this point you conjecture that these (2,3)faceted elements in black, require an inner "mechanical" form, this mod will mean deforming the inner shape, and the elements cannot be simple squares "stuck together", the elements must have extra internal facets, shaped out of solid material, or if the elements aren't solid the surface material must be quite rigid.

Ok so given a kit, being blindfolded and putting the kit together (with unnecessary elements out-of-frame) means you can tell the assembly is "mechanical". What else can you tell, without removing the blindfold, about the structural symmetry? How many "equations of motion" are obvious?

So then, the black plastic cube, with internal slices, is a frame or dynamic structure. It is equivalent to a colored cube that has all the slices stuck together (in "cube formation"), and so has only one "circuit element", which is rotating and inverting the whole structure. This is also equivalent to coloring each facet with a different color (54 for the 3x3x3) which "damps" the color-function - in this case contrast between fixed frequency and intensity of colored squares.

The circuit model is appropriate because even stickering just one facet means you can transport it around the structure. You realise that to track this single element you need to color or mark another part, preferably a fixed part of the structure. All this illustrates that coloring is capacitive, and transport is "induced" when any black element gains a color (a phase change). Thus electronic circuit dynamics is a working model of the structure.

In the case of black plastic, this is like "silicon", and colored plastic is "charge".

Conjecture #2: Only four colors are required if the black faces left are opposites, so an open box with colored sides is constructed.

This one is obvious. As is the restriction on classes of motion - there is no reflection in 3-space because you would need to turn the cube inside out. The only possibility is an optical mirror which reflects an image directly back from a surface, preserving parity. This leaves rotation and translation, which together invert the elements - you can invert an element and return it to its "home slot" so parity is preserved - the parity invariance means any such parity inversion applies strictly to pairs of elements

Okey dokey.
I think I've managed to finally see the connection between the puzzles (starting with the Pocket Cube) and Riemann space (or rather a space of functions).

In 1st year calculus you learn about integrals, and how any arbitrary continuous curve y = f(x) has an approximate solution as a sum of terms that correspond to small width rectangles along the x axis, lying on the curve above.

Riemann surmised that these rectangles can have an arbitrary "fit" and so, they can be an arbitrary width also, and this won't affect a sum if there are enough terms, each an arbitrary width, and arbitrarily close to the curve at the "tall" end. This is called a Riemann integral or Riemann sum, over a continuous(ly differentiable) curve.

The only way to get anything like this is to enumerate all the possible positions a Rubik's puzzle can be in, or, determine the height of each place the function that "fills" each one, for each n, is by some formula that computes the height, and the total number of positions at that value (place) for n.

So, since there is an enumerated table, for n and this includes a second series (of terms) enumerated by a "single step" instruction over all of the instruction set = {URBDLR} and inverse set {U'R'B'D'L'R'}, which we know are "in there". The full set (f(n)) includes the squared moves U[sup]2[/sup], ... which are the "computational identity", since inverting these returns the space to where it started. Thus a squared square move is equal to a "test" or "NOP" kind of cycle, since nothing is changed by an instruction that goes: X[sup]4[/sup].

In fact, this means that any X is an automorphism/endomorphism (it comes as a built-in function of the Pocket Cube and higher-N versions) of the imaginary root of -1, or i.

So the (imaginary) function "square(X)" means we have something like: square: X -> XX = X[sup]2[/sup]; square: square(X) -> square(XX) = X[sup]4[/sup] = 1 (as in, one geometric cube that looks symmetric, none of the "layers" are skewed);



A sequence of Riemann sums. The numbers in the upper right are the areas of the grey rectangles. They converge to the integral of the function.


So the conjecture here is, that the Riemann sums accumulate the positions in each place, which is an nth interval as in the above figure. The final shape appears when you calculate each height (at once, with a recurrence formula, or one-at-a-time using the FTM and QTM, and "processes" U,R, R'U, etc etc).